Problem 69 - Precalculus : Building Concepts and Connections

Question

In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=4 x-1 ; g(x)=\frac{x+1}{4}$$

Step-by-Step Solution

Answer

The composition functions \(f(g(x))\) and \(g(f(x))\) result in \(x\) for both, and their domains are all real numbers as expressed by the interval \(-\infty, \infty\).

Step 1: Find the function composition \(f \circ g\)

To find the function \(f \circ g\), replace all occurrences of \(x\) in \(f(x)\) with \(g(x)\). So, \(f(g(x))=4\left(\frac{x+1}{4}\right)-1\). Simplify this expression to get \(f(g(x))=x+1-1=x\)

Step 2: Find the domain of the function composition \(f \circ g\)

The domain of \(f \circ g\) is all values of \(x\) for which \(g(x)\) is defined and the resulting value \(g(x)\) is in the domain of \(f\). Since \(g(x)=\frac{x+1}{4}\) is defined for all real numbers and all real numbers are in the domain of \(f(x)=4 x-1\), the domain of \(f \circ g\) is \(-\infty, \infty\).

Step 3: Find the function composition \(g \circ f\)

To find the function \(g \circ f\), replace all occurrences of \(x\) in \(g(x)\) with \(f(x)\). So, \(g(f(x))=\frac{4x-1+1}{4}\). Simplify this expression to get \(g(f(x))=x\).

Step 4: Find the domain of the function composition \(g \circ f\)

The domain of \(g \circ f\) is all values of \(x\) for which \(f(x)\) is defined and the resulting value \(f(x)\) is in the domain of \(g\). Since \(f(x)=4 x-1\) is defined for all real numbers and all real numbers are in the domain of \(g(x)=\frac{x+1}{4}\), so the domain of \(g \circ f\) is \(-\infty, \infty\).

Key Concepts

Domain of a Function

The domain of a function represents the set of all possible input values for which the function is defined. It's essential to determine the domain to understand the limitations and behavior of the function. For example, if a function includes a denominator, we must exclude any values that would make the denominator zero, as division by zero is undefined.

When analyzing the domain in a function composition, such as \(f \circ g\), we must consider both the domain of \(g\) and whether the outputs of \(g\) fall within the domain of \(f\). If \(g(x) = \frac{x+1}{4}\), the function is defined for all real numbers because there is no restriction (like division by zero or square roots of negative numbers). The composite function \(f \circ g\) then maintains this unlimited domain, as \(f\)'s operations do not impose new restrictions.

Composite Functions

A composite function is formed when one function is applied to the result of another function. Notation such as \(f \circ g\) represents the function \(f\) composed with \(g\), which means you apply \(g\) first and then \(f\) to the result. The expression for \(f \circ g(x)\) is obtained by substituting \(g(x)\) into \(f\). In our example, applying \(g\) to \(x\) gives us \(g(x) = \frac{x+1}{4}\), and then applying \(f\) to \(g(x)\) results in \(f(g(x)) = x\), which shows a sort of 'cancelling out' effect, simplifying back to \(x\).

This process can sometimes lead to surprising simplifications, where complex expressions reduce down to much simpler forms. It's a powerful tool in precalculus that helps us do things like solve equations involving function compositions or analyze their properties.

Precalculus

The field of precalculus serves as the bridge between the concrete world of algebra and the more abstract realm of calculus. It encompasses a variety of topics, including function composition and the concept of domains. Precalculus equips students with the necessary skills to tackle calculus, such as understanding how different functions behave and interact, and how to manipulate algebraic expressions to solve problems.

The exercise involving \(f\) and \(g\) is a typical precalculus problem. It challenges students to think about how functions can be combined and requires understanding of how to navigate different operations within functions. As such, students engaged in precalculus coursework will frequently encounter these kinds of exercises, which are designed to enhance their ability to think critically about functions before moving on to the study of calculus.